Question

Consider a damped sho having m= 2kg, b= ln 2 ns/m, and k = 100 n/m. find the time ‘t’ after which energy of oscillations will be reduced to half of the initial value?

Answer 1

To find the time 't' after which the energy of an underdamped harmonic oscillator is halved, you calculate the damping ratio
`(√k/m vs b/2m)`and use the exponential decay formula E(t) = E0e-γt to solve for 't'.

Step-by-step explanation:

The question pertains to a damped harmonic oscillator with given mass m = 2 kg, damping constant b = ln(2) Ns/m, and spring constant k = 100 N/m. To find the time 't' after which the energy of the oscillations is reduced to half of its initial value, we need to determine the damping ratio and see if the system is underdamped, critically damped, or overdamped. This involves comparing the square root of the spring constant over mass
`(√k/m)` to the damping constant over twice the mass (b/2m).

For this particular system, we have
`√k/m` greater than b/2m, indicating underdamping. Since the energy in an underdamped oscillator decreases exponentially over time, E(t) = E0e-γt, where γ = b/2m and E0 is the initial energy, the time t at which energy is halved can be determined by solving 1/2 = e-γt. This yields t = ln(2)/γ.